Problem contribution:
Angle Bisector Theorem
Ethan Gabriel S. Ngo — August 30, 2025
Problem 1 (Source: 2022 IWYMIC Individual A4, Rating: 1600, pr. link, sol. link)
In the diagram below, let P be a point on the circumcircle of the equilateral triangle
△ABC such that P B = 24cm and P C = 8cm. If AP and BC intersect at D, then
find the length, in cm, of P D.
Solution. We want to prove that AP is the angle bisector of ∠BP C. Since ACP B is
cyclic, ∠BP C = 180◦ − ∠BAC = 180◦ − 60◦ = 120◦ .
Because ∠BP A and ∠BCA subtend arc BA, ∠P BA = ∠BCA = 60◦ . As such,
∠AP C = ∠BP C −∠BP A = 120◦ −60◦ = 60◦ . Thus, AP is the angle bisector of ∠BP C.
Now, we want to find the side length of P D. First, we can use the Angle Bisector
= BD
= 3. Let BD = 3x and CD = x. By Law of Cosines on
Theorem to get BP
PC
DC
△BP C, we have
BC 2 = BP 2 + P C 2 − 2(BP )(P C)(cos ∠BP C)
16x2 = 242 + 82 − 2(24)(8)(cos 120◦ )
16x2 = 732
√
Solving for x gives x =
Therefore,
133
AB
. Finally, since △ABD ∼ △CP D, we get CP
= BD
=
2
PD
BP
=
DP
√
133
.
4
√
133
4
√
√
3 133
133
2
=
DP
4
DP = 6.
Commentary. Noticing that A, C, P, B are concyclic motivates us to use an angle chase
on △BP C. In particular, since BACP is a cyclic quadrilateral, we can angle chase using
circles to prove that AP is the angle bisector of ∠BP C. This allows us to chase lengths
using Law of Cosines on △BP C and similarity between △BDA and △P DC.
The insight that AP is the angle bisector of ∠BP C can be quite difficult to spot without
a good angle chase. As such, a rating of 1600 would suffice for this problem; this is
suitable for a mid-AMC10 or early AIME problem.
Volume 4, Submitted for Session 2 Unit 2, Option 1
1
Problem contribution:
Angle Bisector Theorem
Ethan Gabriel S. Ngo — August 30, 2025
Problem 2 (Source: Original, Rating: 1600, pr. link, sol. link)
Let △ABC be a triangle where ∠ACB = 2∠ABC, AC = 8, and BC = 6. Find
the length of AB.
Solution. To find the length of AB, we can take advantage of the condition ∠ACB =
2∠ABC. Let CD be the angle bisector of ∠ACB, where D is the foot of the angle
bisector that lies on AB. Thus, ∠ACB = ∠BCD = ∠ACD and so BD = CD.
= BD
= 34 . Let BD = CD = 3k and AD = 4k.
By the Angle Bisector Theorem, BC
AC
AD
In this step, we wish to solve for the value of k.
AC
Since △ADC ∼ △ACB, we have DC
= AD
= AB
. Thus, we obtain (AD)(AB) = AC 2 ,
CB
AC
√
which gives (4k)(7k) = 28k 2 = 64. Solving for k gives k = √47 ; thus, AB = 7k = 4 7.
Commentary. This is a problem in which one must use a certain angle condition that
hints towards drawing an angle bisector (Main Learning #3). Since we know that
∠ACB = 2∠ABC, we can draw an angle bisector that bisects ∠ACB.
We can apply this main learning as a way to angle chase on △ADC and △ABC and
look for similar triangles. With this insight, we can easily solve for the value of AB.
This is suitable for the Angle Bisector Theorem module; the angle condition ∠ACB =
2∠ABC disguises the use of ABT well. As such, a suitable rating for this problem would
be 1600-1750. However, since no heavy computation is needed, and since we can easilly
spot similar triangles from drawing the angle bisector, the approachability of this problem
using basic techniques brings it to 1650.
Volume 4, Submitted for Session 2 Unit 2, Option 1
2